Question

You need to make a benzoic acid-based buffer that has a ph of 5.10. The volume should be 500ml. What masses of the acid and its conjugate base should you use? (Hint: you should not go over 1M in concentration for either acid or conjugate base) Show all Work.

Part B. The pH of the above buffer solution should not go above 5.15. If 10ml of 0.1 M NaOH is added to the buffer, will the pH go past the maximum value?

Answer 1

(a)

For benzoic acid, pKa = 4.2

Assume that we make a buffer with 1M conc.

So,

[acid] + [conj base] = 1 ---(1)

According to Henderson Hasselbach equation:

pH = pKa + log([conj base]/[acid])

Put the values:

5.1 = 4.2 + log([conj base]/[acid]) ---(2)

Solving (1) and (2) we get:

[acid] = 0.11 M

[conj base] = 0.89 M

Since volume = 0.5 L, so

Moles of acid = Conc*Vol = 0.11*0.5 = 0.055

Moles of conj base = Conc*Vol = 0.89*0.5 = 0.445

So,

Mass of benzoic acid = Moles*MW = 0.055*122.12 = 6.72 g

Mass of sodium benzoate (conj base) = Moles*MW = 0.445*144.1 = 64.12 g

(b)

Moles of NaOH added = 0.1*0.01 = 0.001

When NaOH is added, moles of acid decrease and of conj base increase by this amount.

So, finally,

Moles of acid = 0.055-0.001 = 0.054

Moles of conj base = 0.445+0.001 = 0.446

Using Henderson Hasselbach eqn:

pH = pKa + log(moles of conj base/moles of acid)

Put values:

pH = 4.2 + log(0.446/0.054) = 5.12

So we see that pH does not go past maximum allowed value.